# Confidence interval for normal population having both mean and variance..

I'm dealing with this problem:

Let $$X$$ denote the concentration level of a drug in blood of a certain population of $$90$$ people and suppose $$X \sim \mathcal{N}(\mu,\sigma^2)$$ with $$\mu= 100$$ and $$\sigma^2 = 1200$$.

1)Let $$H_0$$ be the following null hypothesis: $$H_0$$ : the mean drug level of the population is equal to $$95$$. Calculate a bilateral confidence interval with 95% significancy for the mean drug concentration.

2)calculate the critical value $$t^*$$ of a $$t$$ student in the case we want to do a bilateral test for $$H_0$$ at 1% level. Is it possible to reject

My attempts:

1)The formula for calculating the (bilateral) confidence interval is $$(\bar{X}_n - z_{\alpha /2}\frac{\sigma}{\sqrt{n}},\bar{X}_n + z_{\alpha /2}\frac{\sigma}{\sqrt{n}})$$

But in this case I know the exact mean $$\mu$$ for the entire population... does this mean $$\bar{X}_{90} = \mu = 100?$$

So I would proceed saying

$$(100 - z_{0.025}\frac{\sqrt{1200}}{\sqrt{90}},100 + z_{0.025}\frac{\sqrt{1200}}{\sqrt{90}})$$

Does this make sense?

2)Why does the problem talk about $$t$$ student? I have both the real mean and standard deviation....

Thanks for the help!

• On the right general track: Some confusion about terminology (see my Answer) and in (2) you need to say $n=90$ and that $\sigma$ is known. If $\sigma$ is unknown and estimated by sample standard deviation $S,$ then the CI uses Student's t distribution as in my Answer.// My answer assumes $\mu$ is not known. If $\mu$ is known there is no point in making a confidence interval for estimating $\mu.$ // Not sure which points of confusion were evident in the original version of your question, but please try to reconcile differences in terminology and method. // Ask in Comment if confusion remains. Aug 27, 2020 at 21:40
• Thanks for the kind answer. Actually I think the question of the exercise is bad posed.. it clearly says that the drug concentration is normally distributed with mean $\mu = 100$ and $\sigma^2 = 1200$. Stated like this I immediately thought that we had a normal distribution with given mean and variance (especially because of the notation used for mean and variance). But both the parameters are actually referred to the specific sample of 90 people coming from a general population we don’t know. So I guess that in this case $\mu = \bar{X}_n$ and $\sigma^2 = S^2$. But it is a real abuse notation. Aug 28, 2020 at 12:08
• This is also justified because in the second question it is asked something about the t student distribution. So in order to have a t distribution we don’t know the real standard deviation but just the sample standard deviation. Aug 28, 2020 at 12:09

(1) If you don't know $$\mu$$ and do know $$\sigma^2 = 1200, \sigma=34.64,$$ and if you have $$n$$ observations from this normal population, then you can find a $$95\%$$ confidence interval for $$\mu$$ of the type $$\bar X \pm 1.96\sigma/\sqrt{n},$$ where $$\bar X$$ is the sample mean. (Note that this is a confidence level, not a level of significance, which would apply to a z test.)
(2) If you know neither $$\mu$$ nor $$\sigma^2$$ but know that the population is normally distributed, then for a sample of size $$n = 90.$$ you would have the two-sided 95% CI $$\bar X \pm 1.987\,S/\sqrt{90},$$ where $$\bar X$$ is the sample mean, $$S$$ is the sample standard deviation, and $$1.987$$ cuts probability $$0.025 = 2.5\%$$ from the upper tail of Student's t distribution with $$\nu = n - 1 =89$$ degrees of freedom (as found in a printed table of t distributions, or by using software, such as R, as below).
qt(.975, 89)

Note: Some authors might use 1.96 (because the t distribution with 89 DF is nearly normal) or 2 (because it is an integer and 'almost correct') instead of $$1.987.$$